\documentclass[a4paper]{article}
\usepackage{amsmath}
\begin{document}
\title{Sum of Two Radicals}
\author{Amit Yaron}
\date{Nov 21, 2025}
\maketitle
\section*{How to Solve $\sqrt[5]{\frac{6x+5}{81x+81}}+\sqrt[5]{\frac{5x+6}{81x+81}}=1$?}
Let us solve it by substitution:
\[
  a=\frac{6x+5}{81x+81},\ b=\frac{5x+6}{81x+81}
\]
Then,
\[
\begin{cases}
  a+b=1\\a^5+b^5=\frac{11}{81}
\end{cases}
\]
By factorizing $a^5+b^5$, we can find values of $ab$ in addition to the value
of $a+b$ we already know.
\begin{align*}
  a^5+b^5&=\underbrace{(a+b)}_{=1}(a^4-a^3b+a^2b^2-ab^3+b^4)\\
  &=a^4-a^3b+a^2b^2-ab^3+b^4\\
  &=a^4+a^2b^2+b^4-a^3b-ab^3\\
  &=a^4+a^2b^2+b^4-ab(a^2+b^2)\\
  &=a^4+2a^2b^2+b^4-a^2b^2-ab(a^2+2ab+b^2-2ab)\\
  &=(a^2+b^2)^2-a^2b^2-ab\big({\underbrace{(a+b)}_{=1}}^2-2ab)\\
  &=(a^2+b^2)^2-(ab)^2-ab(1-2ab)\\
  &=(a^2+b^2)^2-(ab)^2-ab+2(ab)^2\\
  &=(a^2+b^2)^2+(ab)^2-ab\\
  &=(a^2+2ab+b^2-2ab)+(ab)^2-ab\\
  &={\big((a+b)^2-2ab\big)}^2+(ab)^2-ab\\
  &=(1-2ab)^2+(ab)^2-ab\\
  &=1-4ab+4(ab)^2+(ab)^2-ab\\
  &=1-5ab+5(ab)^2
\end{align*}
Let us recall that $a^5+b^5=\frac{11}{81}$
Then,
\begin{align*}
  &5(ab)^2-5ab+1=\frac{11}{81}\Rightarrow\\
  \Rightarrow &5(ab)^2-5ab+1-\frac{11}{81}=0\Rightarrow\\
  \Rightarrow &5(ab)^2-5ab+\frac{70}{81}=0\Rightarrow\\
  \Rightarrow &405-405ab+70=0\Rightarrow\\
  \Rightarrow &ab=\frac{405\pm\sqrt{405^2-4\cdot405\cdot70}}{810}\Rightarrow\\
  \Rightarrow &ab=\frac{405\pm\sqrt{405(405-280)}}{810}\Rightarrow\\
  \Rightarrow &ab=\frac{405\pm\sqrt{81\cdot5\cdot125}}{810}\Rightarrow\\
  \Rightarrow &ab=\frac{405\pm\sqrt{9^2\cdot5^4}}{810}\Rightarrow\\
  \Rightarrow &ab=\frac{405\pm\sqrt{9^2\cdot25^2}}{810}\Rightarrow\\
  \Rightarrow &ab=\frac{405\pm9\cdot25}{810}\Rightarrow\\
  \Rightarrow &ab=\frac{405\pm225}{810}\Rightarrow\\
  \Rightarrow &ab=\frac{9\pm5}{18}
\end{align*}
Now, we have two cases:
\subsection*{Case 1: $ab=\frac{9+5}{18}=\frac{14}{18}=\frac{7}9$}
and
\[
  a+b=1\Rightarrow b=1-a
\]
Thus,
\begin{align*}
  &a(1-a)=\frac{7}9\Rightarrow\\
  \Rightarrow &a-a^2=\frac{7}9\Rightarrow\\
  \Rightarrow &9a-9a^2=7\Rightarrow\\
  \Rightarrow &9a^2-9a+7=0
\end{align*}
But the discriminant is:
\[
  \Delta=9^2-4\cdot9\cdot7=81-252=-171<0
\]
No real solutions.
\subsection*{Case 2: $ab=\frac{9-5}{18}=\frac{4}{18}=\frac{2}9$}
and 
\[
  a+b=1\Rightarrow b=1-a
\]
Thus,
\begin{align*}
  &a(1-a)=\frac{2}9\Rightarrow\\
  \Rightarrow&a(1-a)=\frac{2}9\Rightarrow\\
  \Rightarrow&a-a^2=\frac{2}9\Rightarrow\\
  \Rightarrow&9a-9a^2=2\Rightarrow\\
  \Rightarrow&9a^2-9a+2=0
\end{align*}
Thus,
\begin{align*}
  a=&\frac{9\pm\sqrt{9^2-4\cdot2\cdot9}}{18}\\
  &=\frac{9\pm{81-72}}{18}=\\
  &=\frac{9\pm\sqrt9}{18}=\\
  &=\frac{9\pm3}{18}
\end{align*}
Time to solve for $x$.
Case 1:
  \begin{align*}
    &a=\frac{9+3}{18}=\frac{12}{18}=\frac23\Rightarrow\\
    \Rightarrow &\sqrt[5]{\frac{6x+5}{81x+81}}=\frac23\Rightarrow\\
    \Rightarrow &\frac{6x+5}{81x+81}=\frac{32}{243}\Rightarrow\\
    \Rightarrow &\frac{6x+5}{x+1}=\frac{32}3\Rightarrow\\
    \Rightarrow &32x+32=18x+15\Rightarrow\\
    \Rightarrow &14x+17=0\Rightarrow\\
    \Rightarrow &x=-\frac{17}{14}
  \end{align*}
Case 2:
  \begin{align*}
    &a=\frac{9-3}{18}=\frac{6}{18}=\frac13\Rightarrow\\
    \Rightarrow &\sqrt[5]{\frac{6x+5}{81x+81}}=\frac13\Rightarrow\\
    \Rightarrow &\frac{6x+5}{81x+81}=\frac{1}{243}\Rightarrow\\
    \Rightarrow &\frac{6x+5}{x+1}=\frac{1}3\Rightarrow\\
    \Rightarrow &x+1=18x+15\Rightarrow\\
    \Rightarrow &17x+14=0\Rightarrow\\
    \Rightarrow &x=-\frac{14}{17}
  \end{align*}
So, we have two solutions:
\[
  x=-\frac{14}{17}
\]
or
\[
  x=-\frac{17}{14}
\]
\end{document}